Dividing existing files into multiple files containing specific fields respectively
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0
down vote
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I have a tab separated file_all with 21 fields.
header f1 f2 f3 f4 ...
name1 a b c d ...
...
I need 20 separate files such that file1 contains
header f1
name1 a
...
and file2 contains
header f2
name1 b
...
I can manually do this by
cut -f1,3 file_all > file1
and so on.
Is there an automated way to do this? Thank you.
loops cut
asked Nov 11 at 8:27
Sumin Kim
828
add a comment |
up vote
0
down vote
favorite
I have a tab separated file_all with 21 fields.
header f1 f2 f3 f4 ...
name1 a b c d ...
...
I need 20 separate files such that file1 contains
header f1
name1 a
...
and file2 contains
header f2
name1 b
...
I can manually do this by
cut -f1,3 file_all > file1
and so on.
Is there an automated way to do this? Thank you.
loops cut
asked Nov 11 at 8:27
Sumin Kim
828
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a tab separated file_all with 21 fields.
header f1 f2 f3 f4 ...
name1 a b c d ...
...
I need 20 separate files such that file1 contains
header f1
name1 a
...
and file2 contains
header f2
name1 b
...
I can manually do this by
cut -f1,3 file_all > file1
and so on.
Is there an automated way to do this? Thank you.
loops cut
asked Nov 11 at 8:27
Sumin Kim
828
I have a tab separated file_all with 21 fields.
header f1 f2 f3 f4 ...
name1 a b c d ...
...
I need 20 separate files such that file1 contains
header f1
name1 a
...
and file2 contains
header f2
name1 b
...
I can manually do this by
cut -f1,3 file_all > file1
and so on.
Is there an automated way to do this? Thank you.
loops cut
loops cut
asked Nov 11 at 8:27
Sumin Kim
828
asked Nov 11 at 8:27
Sumin Kim
828
asked Nov 11 at 8:27
Sumin Kim
828
asked Nov 11 at 8:27
Sumin Kim
828
asked Nov 11 at 8:27
Sumin Kim
828
828
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
#!/bin/bash
for ((n=2; n<22; n++))
{
cut -f 1,"$n" file_all > file${n}
}
This is the code I've used to resolve this.
answered Nov 11 at 8:51
Sumin Kim
828
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
#!/bin/bash
for ((n=2; n<22; n++))
{
cut -f 1,"$n" file_all > file${n}
}
This is the code I've used to resolve this.
answered Nov 11 at 8:51
Sumin Kim
828
add a comment |
up vote
0
down vote
#!/bin/bash
for ((n=2; n<22; n++))
{
cut -f 1,"$n" file_all > file${n}
}
This is the code I've used to resolve this.
answered Nov 11 at 8:51
Sumin Kim
828
add a comment |
up vote
0
down vote
up vote
0
down vote
#!/bin/bash
for ((n=2; n<22; n++))
{
cut -f 1,"$n" file_all > file${n}
}
This is the code I've used to resolve this.
answered Nov 11 at 8:51
Sumin Kim
828
#!/bin/bash
for ((n=2; n<22; n++))
{
cut -f 1,"$n" file_all > file${n}
}
This is the code I've used to resolve this.
answered Nov 11 at 8:51
Sumin Kim
828
answered Nov 11 at 8:51
Sumin Kim
828
answered Nov 11 at 8:51
Sumin Kim
828
answered Nov 11 at 8:51
Sumin Kim
828
828
add a comment |
add a comment |
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